# Cassini projection Cassini projection of the world modeled as a highly oblate ellipsoid with flattening 1:2 (= eccentricity ​32)

The Cassini projection (also sometimes known as the Cassini–Soldner projection or Soldner projection) is a map projection described by César-François Cassini de Thury in 1745. It is the transverse aspect of the equirectangular projection, in that the globe is first rotated so the central meridian becomes the "equator", and then the normal equirectangular projection is applied. Considering the earth as a sphere, the projection is composed of the operations:

$x=\arcsin(\cos \varphi \sin \lambda )\qquad y=\arctan \left({\frac {\tan \varphi }{\cos \lambda }}\right).$ where λ is the longitude from the central meridian and φ is the latitude. When programming these equations, the inverse tangent function used is actually the atan2 function, with the first argument sin φ and the second cos φ cos λ.

The reverse operation is composed of the operations:

$\varphi =\arcsin(\sin y\cos x)\qquad \lambda =\operatorname {atan2} (\tan x,\cos y).$ In practice, the projection has always been applied to models of the earth as an ellipsoid, which greatly complicates the mathematical development but is suitable for surveying. Nevertheless, the use of the Cassini projection has largely been superseded by the transverse Mercator projection, at least with central mapping agencies.

## Distortions

Areas along the central meridian, and at right angles to it, are not distorted. Elsewhere, the distortion is largely in a north-south direction, and varies by the square of the distance from the central meridian. As such, the greater the longitudinal extent of the area, the worse the distortion becomes.

Due to this, the Cassini projection works best on long, narrow areas, and worst on wide areas.

## Elliptical form

Cassini is known as a spherical projection, but can be generalised as an elliptical form.

Considering the earth as an ellipse, the projection is composed of these operations:

$N=(1-e^{2}\sin ^{2}\varphi )^{-1/2}$ $T=\tan ^{2}\varphi$ $A=\lambda \cos \varphi$ $C={\frac {e^{2}}{1-e^{2}}}\cos ^{2}\varphi$ $x=N\left(A-T{\frac {A^{3}}{6}}-(8-T+8C)T{\frac {A^{5}}{120}}\right)$ $y=M(\varphi )-M(\varphi _{0})+(N\tan \varphi )\left({\frac {A^{2}}{2}}+(5-T+6C){\frac {A^{4}}{24}}\right)$ and M is the meridional distance function.

The reverse operation is composed of the operations:

$\varphi '=M^{-1}(M(\varphi _{0})+y)$ If $\varphi '={\frac {\pi }{2}}$ then $\varphi =\varphi '$ and $\lambda =0.$ Otherwise calculate T and N as above with $\varphi '$ , and

$R=(1-e^{2})(1-e^{2}\sin ^{2}\varphi ')^{-3/2}$ $D=x/N$ $\varphi =\varphi '-{\frac {N\tan \varphi '}{R}}\left({\frac {D^{2}}{2}}-(1+3T){\frac {D^{4}}{24}}\right)$ $\lambda ={\frac {D-T{\frac {D^{3}}{3}}+(1+3T)T{\frac {D^{5}}{15}}}{\cos \varphi '}}$ 